Is ${647212}$ divisible by $4$ ?
A number is divisible by $4$ if the last two digits are divisible by $4$ . [ Why? We can rewrite the number as a multiple of $100$ plus the last two digits: $ \gray{6472} {12} = \gray{6472} \gray{00} + {12} $ Because $647200$ is a multiple of $100$ , it is also a multiple of $4$ So as long as the value of the last two digits, ${12}$ , is divisible by $4$ , the original number must also be divisible by $4$ Is the value of the last two digits, $12$ , divisible by $4$ Yes, ${12 \div 4 = 3}$, so $647212$ must also be divisible by $4$.